Earth's rotation slows down slightly over time -- How much energy is lost?

[robax25]

Homework Statement

The rotation slows down slightly, about 2.5ms increase for a full rotation after 100 years. Calculate Hoe much rotation energy is lost during 100 years?

Relevant Equations

inertia, I=2/5 mr².

Inertia I = 2/5 mr² m=5.98* 10^24 kg.
r=6.38* 10^6 kg.

I= 9,736 * 10^37 kg.

Earth rotation is V= (2 * pi* R)/T = (2*pi*6.38*10^6m)/(24*3600s)=463 m/s
angular velocity w= V/r =463m/s/(6.38*10^6m)=7.27*10^-5 rad/s
Enerrgy, E= 0.5* I*w²=2.57*10^29 j
I get the same value when period is 24h+2.5ms. How to I calculate energy loss due to Earth rotation slows down.I don't understand how do I get the value of Energy loss? The change of

 

[Steve4Physics]

You have made several mistakes with units and use of standard form (scientific notation). Can you spot them?

The difficulty is that you are dealing with a small difference between two very similar values.

E.g.
x = 1234567890123456789
y = 1234567890123456788
x – y = 1

But if you are trying to work out x – y and you only use x and y to 3 significant figures:
x = 1.23x10¹⁸
y = 1.23x10¹⁸
then x – y will give you zero because of the lack of precision.

The answer is to use some simple calculus.

$E = \frac 1 2 I \omega^2$ (which you used in your method)
and
$\omega = \frac {2\pi}{T}$

Can you find $\frac {dE}{dT}$ and take it from there?

EDITED. Example changed. Errors correct.

 

[robax25]

I get the value -5.959 *10^24 j. Is it lost energy? From my point of view that it is lost of energy. what does it mean that energy is negative?

 

[Steve4Physics]

I get the value 5.959 *10^24 j. Is it lost enegry?

EDIT, Sorry, my mistake. It might be correct - how did you calculate it?

 

[robax25]

yes I get it. The energy that I get after rotation have changed and the lost energy would be 2.56*10^29J. But it is almost same if rotation is 24 hour.If period is 24 h, than the energy is 2.57*10^29j.Lost Energy = (2.57*10^29J)- (5.959*10^24J) =2.5699*10^29j

 

[Steve4Physics]

yes I get it. The energy that I get after rotation have changed and the lost energy would be 2.56*10^29J. But it is almost same if rotation is 24 hour.

Yes, the energy with T = 24 hours and with T = 24hours+2.5ms are very similar. But the difference (energy lost) can be calculated.

Have you read and understood my Post #2?

 

[robax25]

I read it but I have already calculated the value. However, I get the same value. If I miss something, please indicate me.

 

[Steve4Physics]

I read it but I have already calculated the value. However, I get the same value. If I miss something, please indicate me.

EDIT. Sorry I misunderstood and have changed this Post. I will post again when I have looked at your figures more carefully.

 

[Steve4Physics]

Sorry about errors in previous posts.

As far as I can see, you have calculated the energy-loss to be 5.959*10^24J.

You have not shown your working so I can't check. I get a different value - of the order 10^22J.

 

[robax25]

The calculation procedure: If I do derivative of the Energy dE/dt= .5 * I * 4pi^2*(-2)(1/T^3)

I= 9.736* 10^37 kg-m²

dE/dt=.5 * I* 4*pi²*(-2)*(24*3600s)^-3 =-5.959*10^24 J

 

[Steve4Physics]

The calculation procedure: If I do derivative of the Energy dE/dt= .5 * I * 4pi^2*(-2)(1/T^3)

I= 9.736* 10^37 kg-m²

dE/dt=.5 * I* 4*pi²*(-2)*(24*3600s)^-3 =-5.959*10^24 J

Upper case for period. Not dE/dt, but dE/dT.

What units does dE/dT have? (it's not J!)

What does dE/dT mean? Answer: it is the rate of change in energy with respect to the period.

If we consider a very small finite change of period δT, we can write
δE/δT = 5.959*10^24

We can now find the energy-change (δE) resulting from a small period change (δT).

Can you complete the problem? I've got to go out soon so may not be able to reply for a while.

 

[robax25]

should I multiply with 2.5 ms.? Then, I get the result.

 

[Ibix]

I guess the question is, do you understand why multiplying by 2.5ms gets you the answer?

 

[robax25]

because the rate of energy per second.

 

[kuruman]

I guess the question is, do you understand why multiplying by 2.5ms gets you the answer?

I was about to post a full explanation, but in view of this question, I reconsidered and provide a hint instead: Chain rule.

 

[Ibix]

because the rate of energy per second.

Kind of.

The point is that if you imagine drawing a graph of $E$ as a function of $T$ it'll be $E\propto 1/T^2$, right? We can look on the graph and see what energy the Earth has if the period is $T_0$, and we want to know how much that energy changes by if the period increases a small amount $\delta T$. If the graph were a straight line, $E=mT+c$, it'd be easy - the change would be $\left(m(T_0+\delta T)+c\right)-\left(mT_0+c\right)$, which is $m\delta T$. Now our function isn't a straight line, but if $\delta T$ is very small, it's very, very well approximated by one. So for any line, the change would be $\approx m\delta T$. But what's $m$? It's the gradient of the straight line approximating the curve - so it should be the gradient of the curve. So the answer you are looking for is $\frac{dE}{dT}\delta T$, plus some very small correction because the line isn't actually straight.

That's why you multiply by $\delta T$. Keep the trick in mind - for any function $f(x)$ a small change $\delta x$ gives a change $\delta f=\frac{df}{dx}\delta x$, which is further generalised as the chain rule @kuruman mentions.